Question

计算定积分 谢谢

Answer 1

计算过程。

Answer 2

因为xsinx/(2+cos^2x)是偶函数，所以原式=2∫(0,π) xsinx/(2+cos^2x)dx
令t=x-π/2，则x=t+π/2
原式=2∫(-π/2,π/2) (t+π/2)cost/(2+sin^2t)dt
=2∫(-π/2,π/2) tcost/(2+sin^2t)dt+π∫(-π/2,π/2) cost/(2+sin^2t)dt
因为tcost/(2+sin^2t)是奇函数，所以∫(-π/2,π/2) tcost/(2+sin^2t)dt=0
因为cost/(2+sin^2t)是偶函数，
所以∫(-π/2,π/2) cost/(2+sin^2t)dt=2∫(0,π/2) cost/(2+sin^2t)dt
所以原式=2π∫(0,π/2) cost/(2+sin^2t)dt
=2π∫(0,π/2) d(sint)/(2+sin^2t)
=2π*(1/√2)*arctan[(sint)/√2]|(0,π/2)
=√2π*arctan(√2/2)

Answer 3